 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed ${{{v_0}} \over 2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?
A
No
B
Yes, $30^\circ$
C
Yes, $60^\circ$
D
Yes, $45^\circ$

Explanation

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

${{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,$

or $\cos \theta = {1 \over 2}$

$\Rightarrow \cos \theta = \cos 60^\circ$

$\Rightarrow \theta = 60^\circ$
2

AIEEE 2004

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be
A
60 m
B
40 m
C
20 m
D
80 m

Explanation

Assume $a$ be the retardation for both the vehicle then

In case of automobile,

$u_1^2 - 2a{s_1} = 0$

$\Rightarrow u_1^2 = 2a{s_1}$

And in case for car,

$u_2^2 = 2a{s_2}$

$\therefore$ ${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$

$\Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$

$\Rightarrow$ s2 = 80 m
3

AIEEE 2004

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?
A
The velocity vector is tangent to the circle.
B
The acceleration vector is tangent to the circle.
C
The acceleration vector points to the centre of the circle.
D
The velocity and acceleration vectors are perpendicular to each other.

Explanation

Only option $(b)$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.
4

AIEEE 2004

A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to
A
R
B
${1 \over R}$
C
${1 \over {{R^2}}}$
D
${R^2}$

Explanation

Range is same for angle of projection $\theta ,$ and ${90^ \circ } - \theta$

${T_1} = {{2u\sin \theta } \over g},\,\,{T_2} = {{2u\cos \theta } \over g}$

${T_1}{T_2} =$ ${{4{u^2}\sin \theta \cos \theta } \over {{g^2}}}$

= ${2 \over g} \times \left( {{{{u^2}\sin 2\theta } \over g}} \right)$

= ${{2R} \over g}$

(as $R =$${{{{u^2}\sin 2\theta } \over g}}$ )

Hence, ${T_1}{T_2}$ is proportional to $R.$